Naz is swimming the English Channel from France to England. A few minutes ago, she was swimming in calm waters at a velocity (speed and direction) $\vec{v_1}$. The direction of $\vec{v_1}$ is due north, and the speed is $8\,\text{km/h}$. Now, however, she has encountered a current. Without changing the way she swims, she is now moving at a velocity $\vec{v_2}$. The direction of $\vec{v_2}$ is $20^\circ$ west of north, and the speed is $12\,\text{km/h}$. (Assume "due east" is $0^\circ$, "due north" is $90^\circ$, and so on.) What is the speed of the current?
Answer: Consider vector $\vec c$ (depicted below), which represents the current. We can imagine how it would cause Naz to speed up and change direction. It is reasonable to assume that the velocity of the current added with Naz's initial velocity equals her resultant velocity. $\vec c + \vec{v_1} = \vec{v_2}$ Before finding $\vec w$, we'll need to split both vectors into their components. $\vec{v_1}$ is $8\,\text{km/h}$ due north, which can be written as $(0, 8)$. The second vector, $\vec{v_2}$, can be broken into its components using trigonometry. The problem tells us that due north is $90^\circ$, so $20^\circ$ west of north must be $110^\circ$. Since we also know that the magnitude of $\vec{v_2}$ is $12\,\text{km/h}$, we can find the $x$ and $y$ components as follows: (Angle not drawn to scale) $x = 12\cos{(110^\circ)} \approx -4.104$ $y = 12\sin{(110^\circ)} = 11.276$ Therefore, $\vec{v_2} = (-4.104,\, 11.276).$ We can now solve for $\vec c$. $\begin{aligned} \vec c + \vec{v_1} &= \vec{v_2}\\\\ \vec c &= \vec{v_2} - \vec{v_1}\\\\ \vec c &= (-4.104,\, 11.276) - (0, 8)\\\\ \vec c &= (-4.104,\, 3.276) \end{aligned}$ We can find the magnitude of $\vec c$ (i.e., the speed of the current) using the Pythagorean theorem. $\begin{aligned} \| \vec c \|^2 &= (-4.104)^2 + 3.276^2\\\\ \| \vec c \| &= \sqrt{(-4.104)^2 + 3.276^2}\\\\ \| \vec c \| &\approx 5.25 \text{ km/h (rounded to two decimal places)} \end{aligned}$ $\vec c$ is pointing into the second quadrant, so we can find its direction (call it $\theta~$ ) using the arctangent function and adding $180^\circ$. $\begin{aligned} \tan \theta &= \dfrac{3.276}{-4.104}\\\\ \theta &= \arctan{ \left ( -\dfrac{3.276}{4.104} \right ) } \\\\ \theta&\approx -39^\circ \end{aligned}$ Adding $180^\circ$ to this result gives us the actual direction, $141^\circ$. The speed of the current is $5.25 \text{ km}/\text{h}$. The current is flowing in a direction of $141^\circ$.